Problem: Agent Bourne transferred classified files from the CIA mainframe into his flash drive. The drive had $36.5$ megabytes on it before the transfer, and the transfer happened at a constant rate. After $125$ seconds, there were $549$ megabytes on the drive. The drive had a maximum capacity of $1000$ megabytes. What was the speed of the transfer?
Explanation: Let's say that the files were transferred at a rate of $V$ megabytes per second. Then, $V\cdot T$ megabytes were transferred in $T$ seconds. In addition, we know that there were originally $36.5$ megabytes on the drive. The size of the files on the drive is comprised of the files that were on the drive before the transfer began and the files that were transferred. We can express this with the equation $S=36.5+V\cdot T$, where: $S$ represents the size of the files on the drive at a given time (in megabytes) $V$ represents the transfer's rate (in megabytes per second) $T$ represents the time (in seconds) We know that after $125$ seconds $(T={125})$, there were $549$ megabytes on the drive $(S={549})$. Let's plug these values into the equation to find the value of $V$. $ \begin{aligned}{549}&=36.5+V\cdot{125}\\ 125V&=512.5\\ V&=4.1\end{aligned}$ Therefore, the speed of the transfer was $4.1$ megabytes per second. To find how long it took the drive to be completely full, we can plug $S=1000$ into the equation and solve for $T$. $ \begin{aligned}1000&=36.5+4.1T\\ 4.1T&=963.5\\ T&=235\end{aligned}$ The speed of the transfer was $4.1$ megabytes per second. It took the drive $235$ seconds to be completely full.